\newproblem{lay:1_5_11}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 1.5.11}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Describe all solutions of $A\mathbf{x}=\mathbf{0}$ where $A$ is row equivalent to
	\begin{center}
		$B=\begin{pmatrix}1 &-4&-2&0&3&-5\\0&0&1&0&0&-1\\0&0&0&0&1&-4\\0&0&0&0&0&0\end{pmatrix}$
	\end{center}
}
{
  % Solution
	Last equation implies that $x_5-4x_6=0$ or what is the same $x_5=4x_6$. Similarly, the first three equations imply
	\begin{center}
	  $x_1=4x_2+2x_3-3x_5+5x_6$\\
		$x_3=x_6$\\
		$x_5=4x_6$\\
	\end{center}
	Considering the last two equations we may simplify the first equation:
	\begin{center}
	  $x_1=4x_2+2x_6-3(4x_6)+5x_6=4x_2-5x_6$
	\end{center}
	Gathering all this information we deduce that the general solution of $A\mathbf{x}=\mathbf{0}$ is
	\begin{center}
	  $\mathbf{x}=\begin{pmatrix}4x_2-5x_6\\x_2\\x_6\\x_4\\4x_6\\x_6\end{pmatrix}\quad \forall x_2,x_4,x_6\in\mathbb{R}$
	\end{center}
	Note that the free variables ($x_2, x_4, x_6$) are the non-pivot columns in matrix $B$.
}
\useproblem{lay:1_5_11}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
